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kotykmax [81]
2 years ago
9

The figure above shows 4forces 3N, 10N, 3√3N, and 6N acting on a particle P. The resultant of the four forces is.

Physics
1 answer:
Brilliant_brown [7]2 years ago
6 0

As you gave no pic I took them on one lined

  • F_1=3N
  • F_2=10N
  • F_3=3root 3 N
  • F_4=6N

\\ \sf\longmapsto F_{net}=F_1+F_2\dots

\\ \sf\longmapsto F_{net}=3+10+6+3\sqrt{3}

\\ \sf\longmapsto F_{net}=19+3(1.732)

\\ \sf\longmapsto F_{net}=19+5.196

\\ \sf\longmapsto F_{net}=24.196N

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Which is the best definition to describe a transfer of energy?
MissTica

Answer:

a

Explanation:

3 0
2 years ago
A puck of mass 0.110 kg slides across ice in the positive x-direction with a kinetic friction coefficient between the ice and pu
lara [203]

Answer:

a) Ffr = -0.18 N

b) a= -1.64 m/s2

c) t = 9.2 s

d) x = 68.7 m.

e) W= -12.4 J

f) Pavg = -1.35 W

g) Pinst = -0.72 W

Explanation:

a)

  • While the puck slides across ice, the only force acting in the horizontal direction, is the force of kinetic friction.
  • This force is the horizontal component of the contact force, and opposes to the relative movement between the puck and the ice surface, causing it to slow down until it finally comes to a complete stop.
  • So, this force can be written as follows, indicating with the (-) that opposes to the movement of the object.

       F_{frk} = -\mu_{k} * F_{n} (1)

       where μk is the kinetic friction coefficient, and Fn is the normal force.

  • Since the puck is not accelerated in the vertical direction, and there are only two forces acting on it vertically (the normal force Fn, upward, and  the weight Fg, downward), we conclude that both must be equal and opposite each other:

      F_{n} = F_{g} = m*g (2)

  • We can replace (2) in (1), and substituting μk by its value, to find the value of the kinetic friction force, as follows:

       F_{frk} = -\mu_{k} * F_{n} = -0.167*9.8m/s2*0.11kg = -0.18 N (3)

b)

  • According Newton's 2nd Law, the net force acting on the object is equal to its mass times the acceleration.
  • In this case, this net force is the friction force which we have already found in a).
  • Since mass is an scalar, the acceleration must have the same direction as the force, i.e., points to the left.
  • We can write the expression for a as follows:

        a= \frac{F_{frk}}{m} = \frac{-0.18N}{0.11kg} = -1.64 m/s2  (4)

c)

  • Applying the definition of acceleration, choosing t₀ =0, and that the puck comes to rest, so vf=0, we can write the following equation:

        a = \frac{-v_{o} }{t} (5)

  • Replacing by the values of v₀ = 15 m/s, and a = -1.64 m/s2, we can solve for t, as follows:

       t =\frac{-15m/s}{-1.64m/s2} = 9.2 s (6)

d)

  • From (1), (2), and (3) we can conclude that the friction force is constant, which it means that the acceleration is constant too.
  • So, we can use the following kinematic equation in order to find the displacement before coming to rest:

        v_{f} ^{2} - v_{o} ^{2} = 2*a*\Delta x  (7)

  • Since the puck comes to a stop, vf =0.
  • Replacing in (7) the values of v₀ = 15 m/s, and a= -1.64 m/s2, we can solve for the displacement Δx, as follows:

       \Delta x  = \frac{-v_{o}^{2}}{2*a} =\frac{-(15.0m/s)^{2}}{2*(-1.64m/s2} = 68.7 m  (8)

e)

  • The total work done by the friction force on the object , can be obtained in several ways.
  • One of them is just applying the work-energy theorem, that says that the net work done on the object is equal to the change in the kinetic energy of the same object.
  • Since the final kinetic energy is zero (the object stops), the total work done by friction (which is the only force that does work, because the weight and the normal force are perpendicular to the displacement) can be written as follows:

W_{frk} = \Delta K = K_{f} -K_{o} = 0 -\frac{1}{2}*m*v_{o}^{2} =-0.5*0.11*(15.0m/s)^{2}   = -12.4 J  (9)

f)

  • By definition, the average power is the rate of change of the energy delivered to an object (in J) with respect to time.
  • P_{Avg} = \frac{\Delta E}{\Delta t}  (10)
  • If we choose t₀=0, replacing (9) as ΔE, and (6) as Δt, and we can write the following equation:

       P_{Avg} = \frac{\Delta E}{\Delta t} = \frac{-12.4J}{9.2s} = -1.35 W (11)

g)

  • The instantaneous power can be deducted from (10) as W= F*Δx, so we can write P= F*(Δx/Δt) = F*v (dot product)
  • Since F is constant, the instantaneous power when v=4.0 m/s, can be written as follows:

       P_{inst} =- 0.18 N * 4.0m/s = -0.72 W (12)

7 0
2 years ago
Consider a railroad bridge over a highway. A train passing over the bridge dislodges a loose bolt from the bridge, which proceed
photoshop1234 [79]

Answer:

The railroad tracks are 13 m above the windshield (12 m without intermediate rounding).

Explanation:

First, let´s calculate the time it took the driver to travel the 27 m to the point of impact.

The equation for the position of the car is:

x = v · t

Where

x = position at time t

v = velocity

t = time

x = v · t

27 m = 17 m/s · t

27 m / 17 m/s = t

t = 1.6 s

Now let´s calculate the distance traveled by the bolt in that time. Let´s place the origin of the frame of reference at the height of the windshield:

The position of the bolt will be:

y = y0 + 1/2 · g · t²

Where

y = height of the bolt at time t

y0 = initial height of the bolt

g = acceleration due to gravity

t = time

Since the origin of the frame of reference is located at the windshield, at time 1.6 s the height of the bolt will be 0 m (impact on the windshield). Then, we can calculate the initial height of the bolt which is the height of the railroad tracks above the windshield:

y = y0 + 1/2 · g · t²

0 = y0 -1/2 · 9.8 m/s² · (1.6 s)²

y0 = 13 m

8 0
3 years ago
How do metal detcors operate
Masja [62]
Metal detectors work by transmitting an electromagnetic field from the search coil into the ground. Any metal objects (targets) within the electromagnetic field will become energised and retransmit an electromagnetic field of their own.
5 0
2 years ago
Pilots depend on the precision of ------------------------- systems to guide them when they have limited visibility. PLEASE ANSW
kogti [31]

Answer:

Pilots depend on the precision of INSTRUMENT LANDING SYSTEMS to guide them when they have limited visibility.

Explanation:

A pilot is a professionally trained individual that has the ability to control an aircraft following directions from a flight control.

Instrument landing system is a guidance system that makes available an instrument based procedures for guiding the pilot of an aircraft to approach and land safely. This system guides the pilot during unfavorable conditions such as low visibility through the use of radio signals.

There are two types of guidance provided by this system:

-> Lateral guidance subsystem and

-> Vertical guidance subsystem.

Lateral guidance subsystem prevents the aircraft approaching a runway to shift laterally from the path recommended.

Vertical guidance subsystem prevents the aircraft approaching a runway to shift laterally from the recommended path.

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